3.319 \(\int \sec ^8(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=117 \[ \frac{2 i (a+i a \tan (c+d x))^{21/2}}{21 a^7 d}-\frac{12 i (a+i a \tan (c+d x))^{19/2}}{19 a^6 d}+\frac{24 i (a+i a \tan (c+d x))^{17/2}}{17 a^5 d}-\frac{16 i (a+i a \tan (c+d x))^{15/2}}{15 a^4 d} \]

[Out]

(((-16*I)/15)*(a + I*a*Tan[c + d*x])^(15/2))/(a^4*d) + (((24*I)/17)*(a + I*a*Tan[c + d*x])^(17/2))/(a^5*d) - (
((12*I)/19)*(a + I*a*Tan[c + d*x])^(19/2))/(a^6*d) + (((2*I)/21)*(a + I*a*Tan[c + d*x])^(21/2))/(a^7*d)

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Rubi [A]  time = 0.0839321, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3487, 43} \[ \frac{2 i (a+i a \tan (c+d x))^{21/2}}{21 a^7 d}-\frac{12 i (a+i a \tan (c+d x))^{19/2}}{19 a^6 d}+\frac{24 i (a+i a \tan (c+d x))^{17/2}}{17 a^5 d}-\frac{16 i (a+i a \tan (c+d x))^{15/2}}{15 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((-16*I)/15)*(a + I*a*Tan[c + d*x])^(15/2))/(a^4*d) + (((24*I)/17)*(a + I*a*Tan[c + d*x])^(17/2))/(a^5*d) - (
((12*I)/19)*(a + I*a*Tan[c + d*x])^(19/2))/(a^6*d) + (((2*I)/21)*(a + I*a*Tan[c + d*x])^(21/2))/(a^7*d)

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^3 (a+x)^{13/2} \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (8 a^3 (a+x)^{13/2}-12 a^2 (a+x)^{15/2}+6 a (a+x)^{17/2}-(a+x)^{19/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{16 i (a+i a \tan (c+d x))^{15/2}}{15 a^4 d}+\frac{24 i (a+i a \tan (c+d x))^{17/2}}{17 a^5 d}-\frac{12 i (a+i a \tan (c+d x))^{19/2}}{19 a^6 d}+\frac{2 i (a+i a \tan (c+d x))^{21/2}}{21 a^7 d}\\ \end{align*}

Mathematica [A]  time = 1.69881, size = 113, normalized size = 0.97 \[ -\frac{2 a^3 \sec ^9(c+d x) \sqrt{a+i a \tan (c+d x)} (\cos (7 c+10 d x)+i \sin (7 c+10 d x)) (4554 i \cos (2 (c+d x))+630 \tan (c+d x)+2245 \sin (3 (c+d x)) \sec (c+d x)-1311 i)}{33915 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(-2*a^3*Sec[c + d*x]^9*(Cos[7*c + 10*d*x] + I*Sin[7*c + 10*d*x])*(-1311*I + (4554*I)*Cos[2*(c + d*x)] + 2245*S
ec[c + d*x]*Sin[3*(c + d*x)] + 630*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(33915*d*(Cos[d*x] + I*Sin[d*x])^
3)

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Maple [A]  time = 47.629, size = 181, normalized size = 1.6 \begin{align*} -{\frac{2\,{a}^{3} \left ( 8192\,i \left ( \cos \left ( dx+c \right ) \right ) ^{10}-8192\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{9}+1024\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}-5120\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}+448\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}-4032\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +264\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}-3432\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -8300\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+5440\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1615\,i \right ) }{33915\,d \left ( \cos \left ( dx+c \right ) \right ) ^{10}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

-2/33915/d*a^3*(8192*I*cos(d*x+c)^10-8192*sin(d*x+c)*cos(d*x+c)^9+1024*I*cos(d*x+c)^8-5120*sin(d*x+c)*cos(d*x+
c)^7+448*I*cos(d*x+c)^6-4032*cos(d*x+c)^5*sin(d*x+c)+264*I*cos(d*x+c)^4-3432*cos(d*x+c)^3*sin(d*x+c)-8300*I*co
s(d*x+c)^2+5440*cos(d*x+c)*sin(d*x+c)+1615*I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^10

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Maxima [A]  time = 1.10873, size = 103, normalized size = 0.88 \begin{align*} \frac{2 i \,{\left (1615 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{21}{2}} - 10710 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{19}{2}} a + 23940 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{17}{2}} a^{2} - 18088 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{15}{2}} a^{3}\right )}}{33915 \, a^{7} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2/33915*I*(1615*(I*a*tan(d*x + c) + a)^(21/2) - 10710*(I*a*tan(d*x + c) + a)^(19/2)*a + 23940*(I*a*tan(d*x + c
) + a)^(17/2)*a^2 - 18088*(I*a*tan(d*x + c) + a)^(15/2)*a^3)/(a^7*d)

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Fricas [B]  time = 2.32793, size = 682, normalized size = 5.83 \begin{align*} \frac{\sqrt{2}{\left (-32768 i \, a^{3} e^{\left (20 i \, d x + 20 i \, c\right )} - 344064 i \, a^{3} e^{\left (18 i \, d x + 18 i \, c\right )} - 1634304 i \, a^{3} e^{\left (16 i \, d x + 16 i \, c\right )} - 4630528 i \, a^{3} e^{\left (14 i \, d x + 14 i \, c\right )}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}}{33915 \,{\left (d e^{\left (20 i \, d x + 20 i \, c\right )} + 10 \, d e^{\left (18 i \, d x + 18 i \, c\right )} + 45 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 120 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 210 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 252 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 210 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 120 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 45 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/33915*sqrt(2)*(-32768*I*a^3*e^(20*I*d*x + 20*I*c) - 344064*I*a^3*e^(18*I*d*x + 18*I*c) - 1634304*I*a^3*e^(16
*I*d*x + 16*I*c) - 4630528*I*a^3*e^(14*I*d*x + 14*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)/(d*e
^(20*I*d*x + 20*I*c) + 10*d*e^(18*I*d*x + 18*I*c) + 45*d*e^(16*I*d*x + 16*I*c) + 120*d*e^(14*I*d*x + 14*I*c) +
210*d*e^(12*I*d*x + 12*I*c) + 252*d*e^(10*I*d*x + 10*I*c) + 210*d*e^(8*I*d*x + 8*I*c) + 120*d*e^(6*I*d*x + 6*
I*c) + 45*d*e^(4*I*d*x + 4*I*c) + 10*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8*(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \sec \left (d x + c\right )^{8}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(7/2)*sec(d*x + c)^8, x)